Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers. Use sequential hints if required.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

Integer – AIME I, 1993


Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

  • is 107
  • is 870
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Algebra

Check the Answer


Answer: is 870.

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

Try with Hints


Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

taking first two solutions a<b<c<d<500

or,\(1 \leq c-93, c+1 \leq 499\)

or, \(94 \leq c \leq 498 \) gives 405 solutions

and \(1 \leq c-31, c+3 \leq 499\)

or, \(32 \leq c \leq 496\) gives 465 solutions

or, 405+465=870 solutions.

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *